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Advances in Applied Mathematics 46 (2011) 536–562

Contents lists available at ScienceDirect

Advances in Applied Mathematics

www.elsevier.com/locate/yaama

Eulerian quasisymmetric functions and cyclic sieving

Bruce Sagan a, John Shareshian b,∗,1, Michelle L. Wachs c,2

a Department of Mathematics, Michigan State University, East Lansing, MI 48824, United Statesb Department of Mathematics, Washington University, St. Louis, MO 63130, United Statesc Department of Mathematics, University of Miami, Coral Gables, FL 33124, United States

a r t i c l e i n f o a b s t r a c t

Article history:Available online 25 February 2011

Dedicated to Dennis Stanton

MSC:05A1505A3005E0505E18

Keywords:Cyclic sievingMajor indexExcedanceQuasisymmetric functions

It is shown that a refined version of a q-analogue of the Euleriannumbers together with the action, by conjugation, of the subgroupof the symmetric group Sn generated by the n-cycle (1,2, . . . ,n)

on the set of permutations of fixed cycle type and fixed number ofexcedances provides an instance of the cyclic sieving phenomenonof Reiner, Stanton and White. The main tool is a class of symmetricfunctions recently introduced in work of two of the authors.

© 2010 Elsevier Inc. All rights reserved.

1. Introduction

In [15,16], certain quasisymmetric functions, called “Eulerian quasisymmetric functions” are intro-duced and shown to be in fact symmetric functions. These symmetric functions have been useful inthe study of the joint distribution of two permutation statistics, major index and excedance number.There are various versions of the Eulerian quasisymmetric functions. They are defined by first asso-ciating a fundamental quasisymmetric function with each permutation in the symmetric group Sn

and then summing these fundamental quasisymmetric functions over permutations in certain subsets

* Corresponding author.E-mail addresses: [email protected] (B. Sagan), [email protected] (J. Shareshian), [email protected]

(M.L. Wachs).1 Supported in part by NSF Grants DMS 0604233 and 0902142.2 Supported in part by NSF Grants DMS 0604562 and 0902323.

0196-8858/$ – see front matter © 2010 Elsevier Inc. All rights reserved.doi:10.1016/j.aam.2010.01.013


B. Sagan et al. / Advances in Applied Mathematics 46 (2011) 536–562 537

of Sn . To obtain the most refined versions, the cycle-type Eulerian quasisymmetric functions Q λ, j ,one sums the fundamental quasisymmetric functions associated with the permutations having exactlyj excedances and cycle type λ. By summing over all the permutations in Sn having j excedancesand k fixed points, one obtains the less refined version Q n, j,k . The precise definition of the Eulerianquasisymmetric functions and other key terms can be found in Section 2.

Shareshian and Wachs [15,16] derive a formula for the generating function of Q n, j,k which spe-cializes to a (q, r)-analog of a classical formula for the exponential generating function of the Eulerianpolynomials. The (q, r)-analogue of the classical formula is given by

1 +∑n�1

Amaj,exc,fixn (q, t, r)

zn

[n]q! = (1 − tq)expq(rz)

expq(ztq) − tq expq(z), (1)

where

Amaj,exc,fixn (q, t, r) :=

∑σ∈Sn

qmaj(σ )texc(σ )rfix(σ ),

and

expq(z) :=∑n�0

zn

[n]q! .

The cycle-type Eulerian quasisymmetric functions Q λ, j remain somewhat mysterious, and onemight expect that better understanding of them will lead to further results on permutation statistics.In this paper, we provide evidence that this expectation is reasonable. With Theorem 4.1, we prove aconjecture from [16], describing the expansion of Q λ, j , for λ = (n), in terms of the power sum basisfor the space of symmetric functions. Combining Theorem 4.1 with a technique of Désarménien [4],we are able to evaluate, at all nth roots of unity, the cycle-type q-Eulerian numbers

aλ, j(q) :=∑

σ∈Sλ, j

qmaj(σ )−exc(σ ), (2)

where Sλ, j is the set of all σ ∈ Sn having exactly j excedances and cycle type λ. This and an analysisof the excedance statistic on the centralizers C Sn (τ ) of certain permutations τ ∈ Sn enable us toestablish the relationship between the polynomials aλ, j(q) and the cyclic sieving phenomenon ofReiner, Stanton and White [11] given in Theorem 1.2 below.

Notation 1.1. For a positive integer d, ωd will denote throughout this paper an arbitrary complexprimitive dth root of 1.

Theorem 1.2. Let γn = (1,2, . . . ,n) ∈ Sn and let Gn = 〈γn〉 � Sn. Then for all partitions λ of n and j ∈{0,1, . . . ,n − 1}, the group Gn acts on Sλ, j by conjugation and the triple (Gn, Sλ, j,aλ, j(q)) exhibits the cyclicsieving phenomenon. In other words, if τ ∈ Gn has order d then

aλ, j(ωd) = ∣∣C Sn (τ ) ∩ Sλ, j∣∣. (3)

For λ a partition of n, let Sλ be the set of all σ ∈ Sn of cycle type λ and define the cycle typeEulerian polynomial associated with λ as

Amaj,excλ (q, t) :=

∑σ∈Sλ

qmaj(σ )texc(σ ).


538 B. Sagan et al. / Advances in Applied Mathematics 46 (2011) 536–562

Then (3) can be rewritten as

Amaj,excλ

(ωd, tω−1

d

) =∑

σ∈C Sn (τ )∩Sλ

texc(σ ), (4)

which is clearly a refinement of

Amaj,exc,fixn

(ωd, tω−1

d , s) =

∑σ∈C Sn (τ )

texc(σ )sfix(σ ). (5)

Theorem 6.1 below says that both sides of (5) are, in fact, equal to

Amaj,exc,fixnd

(1, t,

sd + t[d − 1]t

[d]t

)[d]

ndt , (6)

which by setting s = 1 yields

Amaj,excn

(ωd, tω−1

d

) = A nd(t) [d]

ndt ,

for all divisors d of n. For cycle-type Eulerian polynomials and all divisors d of n, we obtain in Theo-rem 6.3 the similar looking results,

Amaj,exc(n)

(ωd, tω−1

d

) = (t A n

d −1(t) [d]ndt

)d, (7)

and

Amaj,exc(n+1)

(ωd, tω−1

d

) = t A nd(t) [d]

ndt , (8)

where ( f (t))d is defined in (18) for each polynomial f (t) and positive integer d.The paper is organized as follows. In Section 2, we review definitions of various terms such as

cyclic sieving and Eulerian quasisymmetric functions. We also present some preliminary results on Eu-lerian quasisymmetric functions from [16]. In Section 3 we describe a technique that uses symmetricfunction theory to evaluate certain polynomials at roots of unity based on work of Désarménien [4].Theorem 4.1 mentioned above is proved in Section 4 by means of results of [16] which enable oneto express the cycle-type Eulerian quasisymmetric functions in terms of the less refined version ofEulerian quasisymmetric functions. The proof of Theorem 1.2 appears in Section 5. In Section 6, weprove that both sides of (5) are equal to (6), that (7) and (8) hold, and that another triple exhibits thecyclic sieving phenomenon, namely (Gn, Sn, j,a(n+1), j+1(q)), where Sn, j is the set of all permutationsin Sn with j excedances.

2. Definitions, known facts and preliminary results

2.1. Cyclic sieving

Let G be a finite cyclic group acting on a set X , and let f (q) be a polynomial in q with nonnegativeinteger coefficients. For g ∈ G , let Fix(g) be the set of fixed points of g in X . The triple (G, X, f (q))

exhibits the cyclic sieving phenomenon of Reiner, Stanton and White [11] if for each g ∈ G we have

f (ω|g|) = ∣∣Fix(g)∣∣, (9)

where |g| is the order of g .



Remark 2.1. Since all elements of order d in a cyclic group G generate the same subgroup, they havethe same set of fixed points in any action. Thus our formulation of the cyclic sieving phenomenon isequivalent to the definition given in [11].

Note that if (G, X, f (q)) exhibits the cyclic sieving phenomenon then f (1) = |X |, and interestingexamples arise where f (q) is the generating function for some natural statistic on X , that is, thereexists some useful function s : X → N such that

f (q) =∑x∈X

qs(x).

See [11] and [6] for many examples. Recent work on this subject appears in [10,5,1,2,12,8,9].

2.2. Permutation statistics

Recall that for a permutation σ ∈ Sn acting from the right on [n] := {1, . . . ,n}, the excedance set ofσ is

Exc(σ ) := {i ∈ [n − 1]: iσ > i

}and the descent set of σ is

Des(σ ) := {i ∈ [n − 1]: iσ > (i + 1)σ

}.

The major index of σ is

maj(σ ) :=∑

i∈Des(σ )

i,

and the excedance and descent statistics of σ are, respectively,

exc(σ ) := ∣∣Exc(σ )∣∣,

and

des(σ ) := ∣∣Des(σ )∣∣.

Let Fix(σ ) denote the set of fixed points of σ , that is

Fix(σ ) := {i ∈ [n]: iσ = i

}and let

fix(σ ) := ∣∣Fix(σ )∣∣.

The excedance and descent statistics are equidistributed, and for positive integer n, the nth Eulerianpolynomial An(t) can be defined as

∑σ∈Sn

texc(σ ) = An(t) =∑σ∈Sn

tdes(σ ).



For convenience we set

A0(t) := t−1.

The Eulerian polynomial is also the generating polynomial for the ascent statistic on Sn , asc(σ ) :=|{i ∈ [n − 1]: iσ < (i + 1)σ }|.

For permutation statistics s1, . . . , sk and a positive integer n, define the polynomial

As1,...,skn (t1, . . . , tk) :=

∑σ∈Sn

ts1(σ )1 ts2(σ )

2 · · · tsk(σ )

k .

2.3. Partitions and symmetric functions

We use standard notation for partitions and symmetric functions. References for basic facts are [7,18,13]. In particular, pλ and hλ will denote, respectively, the power sum and complete homogeneoussymmetric functions associated to a partition λ. We use l(λ) to denote the number of (nonzero) partsof λ and m j(λ) to denote the number of parts of λ equal to j. We write Par(n) for the set of allpartitions of n. For λ ∈ Par(n), define the number

zλ :=n∏

j=1

jm j(λ)m j(λ)!.

We use two standard methods to describe a partition λ ∈ Par(n). The first is to write λ =(λ1, . . . , λl(λ)), listing the (nonzero) parts of λ so that λi � λi+1 for all i. The second is to writeλ = 1m1(λ) . . .nmn(λ) , usually suppressing those symbols imi(λ) such that mi(λ) = 0 and writing i1

as simply i. In particular, if n = dk then dk represents the partition with k parts of size d andno other parts. If λ = (λ1, . . . , λk) ∈ Par(n) and q ∈ Q with qλi ∈ P for all i ∈ [k], we write qλ for(qλ1, . . . ,qλk) ∈ Par(qn).

For each σ ∈ Sn , let λ(σ ) denote the cycle type of σ . Given λ ∈ Par(n), we write Sλ for the setof all σ ∈ Sn having cycle type λ. As in (2), we write Sλ, j for the set of those σ ∈ Sλ satisfyingexc(σ ) = j.

For symmetric functions f , g with coefficients in Q[t], f [g] will denote the plethysm of g by f .The same notation will be used for plethysm of symmetric power series with no bound on theirdegree. One such power series is H := ∑

n�1 hn . If we set

L :=∑d�1

μ(d)

dlog(1 + pd)

=∑d�1

μ(d)

d

∑i�1

(−1)i−1

ipi

d,

where μ is the classical Möbius function, then H and L are plethystic inverses, that is,

L[H[ f ]] = H

[L[ f ]] = f (10)

for all symmetric power series f . (This is due to Cadogan, see [3] or [18, Exercise 7.88e].) Note alsothat for any power series h(t, x1, x2, . . .) with coefficients in Q that is symmetric in x1, x2, . . . and anyd ∈ P, we have

pd[h] = h(td, xd

1, xd2, . . .

). (11)

We shall use without further mention the facts ( f + g)[h] = f [h] + g[h] and ( f g)[h] = f [h]g[h].



2.4. q-Analogues

We use the standard notation for polynomial analogues of positive integers, that is, for a positiveinteger n and a variable q, we define

[n]q :=n−1∑j=0

q j = 1 − qn

1 − q

and

[n]q! := [n]q[n − 1]q · · · [1]q.

Also define

[0]q! := 1.

It is well known that for any sequence (k1, . . . ,km) of nonnegative integers whose sum is n, theq-multinomial coefficient [

nk1, . . . ,km

]q:= [n]q!

[k1]q! · · · [km]q!

is always a polynomial in N[q]. The following q-analogue of the multinomial version of the Pascalrecurrence relation is also well known (see [17, (17b)]):

[n

k1, . . . ,km

]q=

m∑i=1

qki+1+···+km

[n − 1

k1, . . . ,ki − 1, . . . ,km

]q, (12)

where (k1, . . . ,km) is a sequence of positive integers whose sum is n. We will need the followingelementary fact, which also plays a role in the work of Reiner, Stanton and White [11].

Proposition 2.2. (See [11, Eq. (4.5)].) Let (k1, . . . ,km) be a sequence of nonnegative integers whose sum is n.If d|n then

[n

k1, . . . ,km

]q

∣∣∣∣q=ωd

=⎧⎨⎩

( nd

k1d ,..., km

d

)if d|ki ∀i ∈ [m],

0 otherwise.

2.5. The Eulerian quasisymmetric functions

Given a permutation σ ∈ Sn , we write σ in one line notation,

σ = σ1 . . . σn,

where σi = iσ . Set

[n] := {ı: i ∈ [n]},

and let w(σ ) be the word in the alphabet A := [n] ∪ [n] obtained from σ by replacing σi with σiwhenever i ∈ Exc(σ ). Order A by



1 < · · · < n < 1 < · · · < n,

and for any word w := w1 . . . wn from A, set

Des(w) := {i ∈ [n − 1]: wi > wi+1

}.

Now, for σ ∈ Sn , define

Dex(σ ) := Des(

w(σ )).

For example, if σ = 641532 then w(σ ) = 641532 and Dex(σ ) = {1,3,5}.Recall now that a quasisymmetric function is a power series (with rational coefficients) f of

bounded degree in variables x1, x2, . . . such that if j1 < · · · < jk and l1 < · · · < lk , then for all a1, . . . ,ak

the coefficients in f of∏k

i=1 xaiji

and∏k

i=1 xaili

are equal. The usual addition, multiplication and scalarmultiplication make the set Q of quasisymmetric functions a Q-algebra that strictly contains thealgebra of symmetric functions. For n ∈ P and S ⊆ [n − 1], set

MonS :={

n∏i=1

x ji : ji � ji+1 for i ∈ [n − 1] and ji > ji+1 if i ∈ S

},

and define the fundamental quasisymmetric function associated with S to be

F S :=∑

x∈MonS

x ∈ Q.

Recall from above that we have defined Sλ, j to be the set of all permutations of cycle type λ with jexcedances. The Eulerian quasisymmetric function associated to the pair (λ, j) is

Q λ, j :=∑

σ∈Sλ, j

FDex(σ ) ∈ Q.

The Eulerian quasisymmetric functions were introduced in [16] as a tool for studying the (maj,exc)q-analogue of the Eulerian polynomials. The connection between the Eulerian quasisymmetric func-tions and the q-Eulerian numbers is given in the following proposition. The stable principal specializa-tion ps is a homomorphism from the algebra of quasisymmetric functions Q to the algebra of formalpower series Q[[q]] defined by ps(xi) = qi−1.

Proposition 2.3. (See [16, Eq. (2.13)].)3 For all partitions λ of n and j ∈ {0,1, . . . ,n − 1}, let aλ, j(q) be asin (2). Then

ps(Q λ, j) = aλ, j(q)∏ni=1(1 − qi)

.

3 Eq. (2.13) in [16] has an extra factor of q j because aλ, j(q) is defined there to be the maj enumerator of Sλ, j rather thanthe maj–exc enumerator.



In [16], it is also shown that in fact Q λ, j is always a symmetric function. If one knows Q (n), jfor all n, j, then a fairly compact explicit formula for each Q λ, j can be obtained from Corollary 6.1of [16], which says that for any λ ∈ Par(n),

n−1∑j=0

Q λ, jtj =

n∏i=1

hmi(λ)

[i−1∑l=0

Q (i),ltl

]. (13)

As noted in [16], if we set

Q n, j :=∑

λ∈Par(n)

Q λ, j

for n � 1, and

Q 0,0 = Q (0),0 = 1,

Eq. (13) implies

∑n, j�0

Q n, jtj =

∑n�0

hn

[ ∑i, j�0

Q (i), jtj]

= H

[ ∑i, j�0

Q (i), jtj],

which by (10) is equivalent to

∑n, j�0

Q (n), jtj = L

[ ∑i, j�0

Q i, jtj]. (14)

Proposition 6.6 of [16] gives an explicit formula for Q n, j in terms of the power sum symmetric func-tion basis,

n−1∑j=0

Q n, jtj =

∑ν∈Par(n)

z−1ν Al(ν)(t)

l(ν)∏i=1

[νi]tpν . (15)

By combining (13), (14) and (15) we obtain a formula for each Q λ, j , which will be used in Section 4to prove a conjecture from [16] giving the expansion of Q (n), j in the power sum basis.

3. A symmetric function technique

We describe here a general technique for evaluating polynomials at roots of unity based on atechnique of Désarménien [4]. This technique provides a key step in our proof of Theorem 1.2. Onecan also prove Theorem 1.1 using Springer’s theory of regular elements in place of the technique wegive here. A description of the relevance of Springer’s work to the cyclic sieving phenomenon appearsin [11].

Given a homogeneous symmetric function F of degree n and a partition ν of n, let χ Fν be the

coefficient of z−1ν pν in the expansion of F in terms of the basis {z−1

ν pν : ν ∈ Par(n)} for the space ofhomogeneous symmetric functions of degree n. That is, χ F

ν is uniquely determined by

F =∑

ν∈Par(n)

χ Fν z−1

ν pν .



Although we will not make use of this, we note that if F is the Frobenius characteristic of a classfunction of Sn then χ F

ν is the value of the class function on permutations of cycle type ν . Recall thatps denotes the stable principal specialization defined in Section 2.5.

The following result is implicit in [4].

Proposition 3.1. Suppose f (q) ∈ Q[q] and there exists a homogeneous symmetric function F of degree n withcoefficients in Q such that

f (q) =n∏

i=1

(1 − qi)ps(F ).

Then for all d,k ∈ P such that n ∈ {dk,dk + 1},

f (ωd) = χ Fν ,

where ν = dk or ν = 1dk.

Proof. By expanding F in the power sum basis for the symmetric functions, we have,

f (q) =n∏

i=1

(1 − qi) ∑

μ∈Par(n)

χ Fμz−1

μ ps(pμ)

=∑

μ∈Par(n)

χ Fμz−1

μ

∏ni=1(1 − qi)∏l(μ)

i=1 (1 − qμi ). (16)

It is shown in [4, Proposition 7.2] that for all μ ∈ Par(n),

Tμ(q) :=∏n

i=1(1 − qi)∏l(μ)

i=1 (1 − qμi )

is a polynomial in q whose value at ωd is given by

Tμ(ωd) ={

zμ if μ = dk or μ = 1dk,

0 otherwise.(17)

We include a proof for the sake of completeness. Since

Tμ(q) =[

nμ1, . . . ,μl(μ)

]q

l(μ)∏i=1

μi−1∏j=1

(1 − q j),

we see that Tμ(q) is a polynomial and that if Tμ(ωd) �= 0 then μi � d for all i. Hence, in the casethat n = dk, it follows from Proposition 2.2 that Tμ(ωd) �= 0 only if μi = d for all i. By Proposition 2.2,

Tdk (ωd) = k!(

d−1∏j=1

(1 − ω

jd

))k

= k!dk.



Similarly, in the case that n = dk + 1, we use (12) and Proposition 2.2 to show that Tμ(ωd) equalsk!dk if μ = 1dk and is 0 otherwise. Hence, in either case, (17) holds. Now by plugging (17) into (16)we obtain the desired result. �

We will use Propostion 3.1 to evaluate the cycle-type Eulerian numbers aλ, j(q) at all the mthroots of unity, where m ∈ {n − 1,n}. We see from Proposition 2.3 that we already have the requiredsymmetric function, namely Q λ, j . We thus obtain the first step in our proof of Theorem 1.2.

Proposition 3.2. Let λ ∈ Par(n) and let d,k ∈ P. If dk = n then

aλ, j(ωd) = χQ λ, j

dk ,

and if dk = n − 1 then

aλ, j(ωd) = χQ λ, j

1dk .

In [16] a formula for the coefficients χQ (n), jν is conjectured. This formula turns out to be just what

we need to prove Theorem 1.2. In the next section we present the conjecture and its proof.

Remark 3.3. In [16] it is conjectured that Q λ, j is the Frobenius characteristic of some representationof Sn . A proof of this conjecture was recently obtained in [14]. By Proposition 3.2 and Theorem 1.2,the restriction of the representation in question to Gn is isomorphic to the permutation representationfor the action of Gn on Sλ, j .

4. The expansion of Q (n), j

In this section we present a key result of our paper (Theorem 4.1), which was conjectured in [16].For a power series f (t) = ∑

j�0 a jt j and an integer k, let f (t)k be the power series obtained from

f (t) by erasing all terms a jt j such that gcd( j,k) �= 1, so

f (t)k :=∑

gcd( j,k)=1

a jtj. (18)

For example, if f (t) = t + 3t2 − 5t3 + 7t4 then f (t)2 = t − 5t3.For a partition ν = (ν1, . . . , νk), set

g(ν) := gcd(ν1, . . . , νk).

Theorem 4.1. (See [16], Conjecture 6.5.) For ν = (ν1, . . . , νk) ∈ Par(n), set

Gν(t) :=(

t Ak−1(t)k∏

i=1

[νi]t

)g(ν)

.

Then

n−1∑j=0

Q (n), jtj =

∑ν∈Par(n)

z−1ν Gν(t)pν . (19)



Theorem 4.1 can be restated as follows. Since Q (n), j is a homogeneous symmetric function of de-gree n, it can be expanded in the basis {z−1

λ pλ: λ ∈ Par(n)}. Thus, the theorem says that the expansioncoefficient of z−1

ν pν is 0 if gcd( j, g(ν)) �= 1, while if gcd( j, g(ν)) = 1 then the expansion coefficient

equals the coefficient of t j in t Al(ν)−1(t)∏l(ν)

i=1[νi]t .In order to prove Theorem 4.1 we need two lemmas. As above, we write μ for the classical Möbius

function on P, and recall that

∑d|n

μ(d) ={

1, n = 1,

0, otherwise.(20)

Lemma 4.2. For a partition ν = (ν1, . . . , νl), we have

Gν(t) =∑

d|g(ν)

μ(d)dl−1td Al−1(td) l∏

i=1

[νi

d

]td

. (21)

Proof. It is known (and follows, for example, from [17, Theorem 4.5.14]) that for any positive integerk we have

t Ak−1(t)

(1 − t)k=

∑j�0

jk−1t j. (22)

It follows directly from the definition of f (t)d that for any power series g,h and any d ∈ P wehave

(g(t)h

(td))

d = g(t)dh(td). (23)

We see now that

Gν(t) =(

t Al−1(t)l∏

i=1

1 − tνi

1 − t

)g(ν)

=(

t Al−1(t)

(1 − t)l

l∏i=1

(1 − tνi

))g(ν)

=(∑

j�0

jl−1t j)

g(ν)

l∏i=1

(1 − tνi

)

=∑

j: gcd(g(ν), j)=1

jl−1t jl∏

i=1

(1 − tνi

),

the third equality above following from (22) and (23).Now

∑d|g(ν)

μ(d)∑a�0

(ad)l−1tad =∑j�0

jl−1t j∑

d|gcd( j,g(ν))

μ(d)

=∑

j: gcd( j,g(ν))=1

jl−1t j,



the second equality following from (20). We see now that

Gν(t) =( ∑

d|g(ν)

μ(d)∑a�0

(ad)l−1tad) l∏

i=1

(1 − tνi

)

=( ∑

d|g(ν)

μ(d)dl−1 td Al−1(td)

(1 − td)l

) l∏i=1

(1 − tνi

)

=∑

d|g(ν)

μ(d)dl−1td Al−1(td) l∏

i=1

1 − tνi

1 − td,

the second equality following from (22). �Lemma 4.3. We have

1 +∑k�1

Ak(t)zk

k! = exp

(∑l�1

t Al−1(t)zl

l!)

. (24)

Proof. We apply the exponential formula (see [18, Corollary 5.1.6]) to the Eulerian polynomials. Forany permutation σ in Sn let π(σ ) be the partition of the set [n] whose blocks are the supports ofthe cycles in the cycle decomposition of σ . Let Πn be the set of all partitions of the set [n]. For anypartition π in Πn set

Aπ (t) :=∑σ∈Sn

π(σ )=π

texc(σ ).

Then

An(t) =∑

π∈Πn

Aπ (t),

and

Aπ (t) =k∏

i=1

A{Bi}(t) =k∏

i=1

A(|Bi |)(t),

where π = {B1, . . . , Bk}. It therefore follows from the exponential formula that

1 +∑k�1

Ak(t)zk

k! = exp

(∑l�1

A(l)(t)zl

l!)

.

To complete the proof we observe that

A(l)(t) = t Al−1(t). (25)

Indeed, if l = 1 then both sides of the equation are equal to 1. For l > 1 and σ ∈ S(l) , write σ incycle notation (x1, x2, . . . , xl) with xl = l. Now let v(σ ) = x1 . . . xl−1, a permutation in Sl−1 in one line



notation. The excedance set of σ is the union of {xl−1} and {xi: i is an ascent of v(σ )}. Since v is abijection from S(l) to Sl−1, Eq. (25) holds. �Proof of Theorem 4.1. We have

∑n�1

n−1∑j=0

Q (n), jtj = L

[∑i�1

i−1∑j=0

Q i, jtj

]

= L

[∑k�1

∑ν: l(ν)=k

z−1ν Ak(t)

k∏h=1

[νh]tpνh

]

=∑d�1

μ(d)

d

∑i�1

(−1)i−1

ipi

d

[∑k�1

∑ν: l(ν)=k

z−1ν Ak(t)

k∏h=1

[νh]tpνh

]

=∑d�1

μ(d)

d

∑i�1

(−1)i−1

i

(∑k�1

∑ν: l(ν)=k

z−1ν Ak

(td) k∏

h=1

[νh]td pdνh

)i

=∑d�1

μ(d)

dlog

(1 +

∑k�1

∑ν: l(ν)=k

z−1ν Ak

(td) k∏

h=1

[νh]td pdνh

), (26)

the first equality following from (14), the second from (15), the third from the definition of L and thefourth from (11).

For any k ∈ P, let Mk be the set of all sequences a = (a1,a2, . . .) of nonnegative integers such that∑r�1 ar = k. Then

∑ν: l(ν)=k

z−1ν

k∏h=1

[νh]td pdνh =∑

ν: l(ν)=k

1∏r�1 mr(ν)!

∏r�1

( [r]td pdr

r

)mr(ν)

= 1

k!∑

a∈Mk

(k

a1,a2, . . .

) ∏r�1

( [r]td pdr

r

)ar

= 1

k!(∑

r�1

[r]td pdr

r

)k

. (27)

We see now that

∑n�1

n−1∑j=0

Q (n), jtj =

∑d�1

μ(d)

dlog

(1 +

∑k�1

Ak(td)

k!(∑

r�1

[r]td pdr

r

)k)

=∑d�1

μ(d)

d

∑k�1

td Ak−1(td)

k!(∑

r�1

[r]td pdr

r

)k

=∑d�1

μ(d)

d

∑k�1

∑ν: l(ν)=k

z−1ν td Ak−1

(td) k∏

i=1

[νi]td pdνi



=∑k�1

∑d�1

μ(d)dk−1td Ak−1(td) ∑

ν: l(ν)=k

z−1dν

k∏i=1

[νi]td pdνi

=∑k�1

∑ν: l(ν)=k

z−1ν pν

∑d|g(ν)

μ(d)dk−1td Ak−1(td) k∏

i=1

[νi

d

]td

=∑k�1

∑ν: l(ν)=k

z−1ν pνGν(t).

Indeed, first equality is obtained by combining (26) and (27), the second equality is obtained from(24), the third follows from (27), the fourth and fifth are obtained by straightforward manipulations,and the last follows from (21). �5. The proof of Theorem 1.2

5.1. The expansion coefficients χQ λ, j

dk

To compute the expansion coefficients χQ λ, j

dk , we will need to obtain results like Theorem 4.1with the partition (n) replaced by an arbitrary partition λ, but in such results we will only needthe coefficients of power sum symmetric functions of the form p(d,...,d) . We begin with a definitiongeneralizing that of f (t)d . For a power series f (t) = ∑

j a jt j and positive integers b, c, let f (t)b,c be

the power series obtained from f by erasing all terms aiti such that gcd(i,b) �= c, so

f (t)b,c :=∑

gcd(i,b)=c

aiti .

For example, if f (t) = 1 + 2t + 3t2 + 4t3 + 5t4 then f (t)6,2 = 3t2 + 5t4. For any power series g,h, wehave

(g(t)h

(tb))

b,c = g(t)b,ch(tb). (28)

We will use the following result.

Lemma 5.1. Let k,b, c ∈ P and assume that c|b. Then

(t Ak−1(t)[b]k

t

)b,c = ck−1(tc Ak−1

(tc)[b/c]k

tc

)b,c . (29)

Proof. We have

(t Ak−1(t)[b]k

t

)b,c =

(t Ak−1(t)

(1 − t)k

(1 − tb)k

)b,c

=(

t Ak−1(t)

(1 − t)k

)b,c

(1 − tb)k

= (1 − tb)k ∑

j: gcd( j,b)=c

jk−1t j

= (1 − tb)k ∑

i: gcd(i,b/c)=1

(ic)k−1tic



= ck−1(1 − tb)k ∑i: gcd(i,b/c)=1

ik−1tic

= ck−1(1 − tb)k(

tc Ak−1(tc)

(1 − tc)k

)b,c

= ck−1(

tc Ak−1(tc) (1 − tb)k

(1 − tc)k

)b,c

= ck−1(tc Ak−1(tc)[b/c]k

tc

)b,c.

Indeed, the second and seventh equalities follow from (28), the third and sixth follow from (22), andthe rest are straightforward. �

We begin our computation of χQ λ, j

dk by considering first the case where all parts of λ have thesame size. For λ,ν ∈ Par(n) and j ∈ {0,1, . . . ,n − 1}, set

χλ, jν := χ

Q λ, jν

and

πν :=(

n

ν1, . . . , νl(ν)

)1∏n

j=1 m j(ν)! . (30)

Theorem 5.2. Let n, r,m,d,k ∈ P with n = rm = dk. Set

Par(m;d, r) := {μ = (μ1, . . . ,μl(μ)) ∈ Par(m): μi |d|rμi for all i ∈ [

l(μ)]}

.

Then

n−1∑j=0

χrm, jdk t j =

∑μ∈Par(m;d,r)

π rd μ

l(μ)∏i=1

(t A r

d μi−1(t)[d]rd μit

)d,μi

. (31)

Proof. Note that (13) implies that

n−1∑j=0

Q rm, jtj = hm

[r−1∑j=0

Q (r), jtj

]. (32)

Now

hm

[r−1∑j=0

Q (r), jtj

]=

∑μ∈Par(m)

z−1μ pμ

[r−1∑j=0

Q (r), jtj

]

=∑

μ∈Par(m)

z−1μ

l(μ)∏i=1

pμi

[ ∑ν∈Par(r)

z−1ν Gν(t)pν

]

=∑

μ∈Par(m)

z−1μ

l(μ)∏i=1

( ∑ν∈Par(r)

z−1ν Gν

(tμi

)pμiν

).



Indeed, the first equality follows from the well-known expansion (see any of [7,13,18]) of hm in thepower sum basis, the second from Theorem 4.1 and the third from (11).

On the other hand, it follows from the definition of χλ, jν that

n−1∑j=0

Q rm, jtj =

∑ν∈Par(n)

z−1ν

∑j�0

χrm, jν pνt j.

We see now equating the coefficients of pdk on both sides of (32) yields

n−1∑j=0

χrm, jdk t j = zdk

∑μ∈Par(m;d,r)

z−1μ

l(μ)∏i=1

z−1( dμi

)rμi/d G( dμi

)rμi/d

(tμi

). (33)

Now for all μ ∈ Par(m;d, r), we have

zdk z−1μ

l(μ)∏i=1

z−1( dμi

)rμi/d = dkk!∏mj=1 m j(μ)!∏l(μ)

i=1 μi(rμi

d )!( dμi

)rμi/d

= k!∏l(μ)

i=1 μ(rμi/d)−1i∏m

j=1 m j(μ)!∏l(μ)

i=1 (rμi

d )!

= π rd μ

l(μ)∏i=1

μ(rμi/d)−1i , (34)

and

G( dμi

)rμi/d

(tμi

) = (t A(rμi/d)−1(t)[d/μi]rμi/d

t

)d/μi

∣∣t=tμi

= (tμi A(rμi/d)−1

(tμi

)[d/μi]rμi/dtμi

)d,μi

. (35)

We now have

n−1∑j=0

χrm, jdk t j =

∑μ∈Par(m;d,r)

π rd μ

l(μ)∏i=1

μrd μi−1i

(tμi A r

d μi−1(tμi

)[d/μi]rd μi

tμi

)d,μi

=∑

μ∈Par(m;d,r)

π rd μ

l(μ)∏i=1

(t A r


)d,μi

,

the first equality being obtained by substituting (34) and (35) into (33), and the second followingfrom Lemma 5.1. �

We use Theorem 5.2 to handle general λ.

Theorem 5.3. Say λ ∈ Par(n) and n = kd.

1. If there is some r ∈ [n] such that d does not divide rmr(λ) then χλ, jdk = 0 for all 0 � j � n − 1.



2. If d divides rmr(λ) for all r ∈ [n] then

n−1∑j=0

χλ, jdk t j =

( nd

1m1(λ)d ,

2m2(λ)d , . . . ,

nmn(λ)d

) n∏r=1

rmr(λ)−1∑j=0

χrmr (λ), jdrmr (λ)/d t j.

Proof. It follows directly from (13) that

n−1∑j=0

Q λ, jtj =

n∏r=1

rmr(λ)−1∑j=0

Q rmr (λ), jtj. (36)

Expressing both sides of (36) in terms of the power sum basis, we get

∑μ∈Par(n)

z−1μ

n−1∑j=0

χλ, jμ t jpμ =

n∏r=1

∑ν∈Par(rmr(λ))

z−1ν

rmr(λ)−1∑j=0

χrmr (λ), jν t jpν . (37)

Equating coefficients of pdk in (37) we see that if d does not divide every rmr(λ) then χλ, jdk = 0 for

all j, while if d divides every rmr(λ) then

n−1∑j=0

χλ, jdk t j = zdk

n∏r=1

z−1drmr (λ)/d

rmr(λ)−1∑j=0

χrmr (λ), jdrmr (λ)/d t j

= dn/d(n/d)!∏nr=1 drmr(λ)/d(rmr(λ)/d)!

n∏r=1

rmr(λ)−1∑j=0

χrmr (λ), jdrmr (λ)/d t j

=( n

d1m1(λ)

d ,2m2(λ)

d , . . . ,nmn(λ)

d

) n∏r=1

rmr(λ)−1∑j=0

χrmr (λ), jdrmr (λ)/d t j. �

5.2. The permutation character θλ, j of Gn

Note that, upon considering cycle notation for elements of Sn , it is straightforward to show thatif σ ∈ Sλ, j then γ −1

n σγn ∈ Sλ, j . Thus the claim in Theorem 1.2 that Gn acts on Sλ, j is correct. Letθλ, j denote the permutation character of the action of Gn on Sλ, j . Hence, θλ, j(τ ) is the number of

elements of Sλ, j centralized by τ ∈ Gn . For ν ∈ Par(n), let θλ, jν = θλ, j(τ ), where τ is any permutation

of cycle type ν . Since all τ ∈ Gn have cycle type of the form dk , where dk = n, we need only concernourselves with ν = dk .

With Theorems 5.2 and 5.3 in hand, we now produce matching results for the permutation char-acters θλ, j of Gn . Again we begin with the case where λ = rm for some divisor r of n. Before doingso, we derive, in the form most useful for our arguments, some known facts about centralizers in Sn

of elements of Gn , along with straightforward consequences of these facts.Fix positive integers n,k,d with n = kd. Set τ = γ −k

n ∈ Gn . Note that C Sn (τ ) = C Sn (γk

n ). For i ∈ [n],we have

iτ ={

i − k, i > k,

i − k + n, i � k.



Now τ has cycle type dk , and we can write τ as the product of k d-cycles, τ = τ1 . . . τk , where τi hassupport

Xi := {j ∈ [n]: j ≡ i mod k

}.

It follows that if σ ∈ C Sn (τ ) then for each i ∈ [k] there is some j ∈ [k] such that Xiσ = X j . Thuswe have an action of C Sn (τ ) on {X1, . . . , Xk}, which gives rise to a homomorphism

Φ : C Sn(τ ) → Sk.

Given ρ ∈ Sk , define ρ̂ ∈ Sn to be the element that, for r ∈ [k] and q ∈ {0, . . . ,d − 1}, maps r + qk torρ + qk. It is straightforward to check that ρ̂ ∈ C Sn (τ ) and Φ(ρ̂) = ρ . Moreover, if we set

R := {ρ̂: ρ ∈ Sk},

then R � C Sn (τ ) and the restriction of Φ to R is an isomorphism. It follows that if we set K =kernel(Φ) then C Sn (τ ) is the semidirect product of K and R . Now

K = {σ ∈ C Sn(τ ): Xiσ = Xi for all i ∈ [k]} =

k∏i=1

C S Xi(τi).

Since every d-cycle in Sd generates its own centralizer in Sd , we have

K =k∏

i=1

〈τi〉 ={

k∏i=1

τeii : e1, . . . , ek ∈ {0, . . . ,d − 1}

}.

Now, given ρ ∈ Sk and e1, . . . , ek ∈ {0, . . . ,d − 1}, set

σ := τe11 . . . τ

ekk ρ̂ ∈ C Sn (τ ).

For r ∈ [k] and q ∈ {0, . . . ,d − 1}, we have (with σ acting on the right)

(r + qk)σ ={

rρ + (q − er)k, q � er,

rρ + (q − er)k + n, q < er .(38)

It follows that r + qk ∈ Exc(σ ) if and only if either q < er or er = 0 and r < rρ . We collect in the nextlemma some useful consequences of what we have just seen.

Lemma 5.4. Let n = dk and let τ = γ −kn . Let σ ∈ C Sn (τ ). Then there exist unique ρ ∈ Sk and e1, . . . , ek ∈

{0, . . . ,d − 1} such that

σ = τe11 . . . τ

ekk ρ̂,

and if we define E0 to be the number of r ∈ [k] such that er = 0 and r ∈ Exc(ρ), then

exc(σ ) = dE0 +k∑

i=1

ei . (39)



Note that the unique ρ ∈ Sk of Lemma 5.4 is equal to Φ(σ ) defined above. For μ ∈ Par(k) and anydivisor r of n, set

Cμ := {σ ∈ C Sn (τ ): Φ(σ) ∈ Sμ

}and

Cμ,r := Cμ ∩ Srn/r ,

so Cμ,r consists of those σ ∈ C Sn (τ ) such that σ has cycle type rn/r and Φ(σ ) has cycle type μ.

Lemma 5.5. For any divisor r of n, we have

∑σ∈C(k),r

texc(σ ) ={

(t Ak−1(t)[d]kt )d, n

r, if k|r,

0, otherwise.

Proof. We begin by showing that

C(k) =⊎k|r|n

C(k),r . (40)

Certainly the union on the right side of (40) is contained in C(k) , so we prove that this unioncontains C(k) . Let σ ∈ C(k) . By Lemma 5.4, we have

σ = τe11 . . . τ

ekk ρ̂

for unique ρ ∈ S(k) and e1, . . . , ek ∈ {0, . . . ,d − 1}. It follows from (38) that for each j ∈ [n] we have

jσ k ≡ j − kk∑

i=1

ei mod n. (41)

Moreover, if jσ l ≡ j mod k then k|l. Hence each cycle length in the cycle decomposition of σ is amultiple of k.

We claim that all cycles in the cycle decomposition of σ have length sk, where s is the order ofk∑k

i=1 ei in Zn . Indeed, it follows from (41) that for all j ∈ [n],

jσ sk ≡ j − skk∑

i=1

ei ≡ j mod n,

which implies that jσ sk = j. Hence the order of σ in Sn divides sk. It follows that every cycle lengthin the cycle decomposition of σ divides sk. Now we need only show that sk divides the length ofeach cycle. Suppose α is a cycle of length r and j is an element in the support of α. We have k|rsince k divides the length of every cycle. Again using (41) we have,

j = jσ r = j(σ k)r/k ≡ j − r

kk

k∑i=1

ei mod n,



which implies that (r/k)k∑k

i=1 ei ≡ 0 mod n. Thus s, the order of k∑k

i=1 ei , divides r/k, which im-plies that sk divides r. We have therefore shown that sk divides the length of every cycle, and sincewe have already shown that every cycle length divides sk, we conclude that all cycles in the cycledecomposition of σ have the same length sk, that is, σ ∈ C(k),r for some r satisfying k|r|n, as claimedin (40).

We have also shown that C(k),r = ∅ if k does not divide r. Thus the claim of the lemma holds whenk does not divide r.

Next we show that if σ ∈ C(k),r then

gcd(exc(σ ),d

) = n

r. (42)

As above, write σ = τe11 . . . τ

ekk ρ̂ . Since d|n, it follows from (39) that

gcd(exc(σ ),d

) = gcd

(k∑

i=1

ei,d

). (43)

Since k∑k

i=1 ei has order r/k in (Zn,+), we have that

r

k= n

gcd(n,k∑k

i=1 ei)= d

gcd(d,∑k

i=1 ei),

the first equality following from simple facts about modular arithmetic and the second from the factthat n = dk. Now we have

gcd

(d,

k∑i=1

ei

)= kd

r, (44)

and combining (44) with (43) gives (42).Combining (40) and (42), we get

∑σ∈C(k),r

texc(σ ) =( ∑

σ∈C(k)

texc(σ )

)d, n

r

(45)

for each divisor r of n.For ρ ∈ Sk and i ∈ [k], set

fρ,i(t) :={

t[d]t if i ∈ Exc(ρ),

[d]t otherwise.

Then

∑σ∈Φ−1(ρ)

texc(σ ) =k∏

i=1

fρ,i(t) = texc(ρ)[d]kt ,

the first equality following from Lemma 5.4. It follows now from (25) that



∑σ∈C(k)

texc(σ ) = t Ak−1(t)[d]kt , (46)

and combining (45) and (46) yields the lemma. �Lemma 5.6. For divisors k, r of n and μ ∈ Par(k), we have

∑σ∈Cμ,r

texc(σ ) = πμ

l(μ)∏i=1

∑σ∈C(μi ),r

texc(σ ),

where πμ is defined as in (30).

Proof. Given σ ∈ Cμ,r , we write as usual σ = τe11 . . . τ

ekk ρ̂ . Now ρ ∈ Sk has cycle type μ, so we can

write ρ = ρ1 . . . ρl(μ) as a product of disjoint cycles whose lengths form the partition μ. For i ∈ [l(μ)],let Bi be the support of ρi . We may assume that |Bi | = μi for all i. Set

β(σ ) := {B1, . . . , Bl(μ)},

so β(σ ) is a partition of [k]. For i ∈ [l(μ)], set

σi :=( ∏

j∈Bi

τe j

j

)ρ̂i ∈ Sn.

The supports of both ρ̂i and∏

j∈Biτ

e j

j are contained in

B̂ i := { j + qk: j ∈ Bi, 0 � q � d − 1}.

It follows that ρ̂i and Π j∈Bh τe j

j commute for all i �= h, so

σ =l(μ)∏i=1

σi .

Moreover,

exc(σ ) =l(μ)∑i=1

exc(σi). (47)

For i ∈ [l(μ)], define f i to be the unique order preserving bijection from B̂ i to [dμi], and set

σ i := f −1i σi f i .

Then, for each i ∈ [l(μ)], we have σ i ∈ C(μi),r and

exc(σ i) = exc(σi). (48)



Let Πμ be the set of partitions of [k] that have m j(μ) blocks of size j for each j. For each partitionX ∈ Πμ , set

CX := {σ ∈ Cμ,r: β(σ ) = X

}.

The map from CX to∏l(μ)

i=1 C(μi),r sending σ to (σ 1, . . . , σ l(μ)) is a bijection. Given (47) and (48), wesee that

∑σ∈Cμ,r

texc(σ ) =∑

X∈Πμ

∑σ∈CX

texc(σ )

= |Πμ|l(μ)∏i=1

∑ρ∈C(μi ),r

texc(ρ).

It is straightforward to see that |Πμ| = πμ , so the lemma follows. �Theorem 5.7. Let n, r,m,d,k ∈ P with n = rm = dk. As in Theorem 5.2, let

Par(m;d, r) = {μ = (μ1, . . . ,μl(μ)) ∈ Par(m): μi |d|rμi for all i ∈ [

l(μ)]}

.

Then

n−1∑j=0

θrm, jdk t j =

∑μ∈Par(m;d,r)

π rd μ

l(μ)∏i=1

(t A r


)d,μi

. (49)

Proof. We have

n−1∑j=0

θrm, jdk t j =

∑σ∈C Sn (γ k

n )∩Srm

texc(σ )

=∑

μ∈Par(k)

∑σ∈Cμ,r

texc(σ )

=∑

μ∈Par(k)

πμ

l(μ)∏i=1

∑σ∈C(μi ),r

texc(σ )

=∑

μ∈Par(k;r,d)

πμ

l(μ)∏i=1

(t Aμi−1(t)[d]μi

t

)d,dμi/r .

Indeed, the first two equalities follow immediately from the definitions of θrm, jdk and Cμ,r , respectively,

while the third follows from Lemma 5.6 and the fourth from Lemma 5.5.Now for μ ∈ Par(k; r,d), set ν := ν(μ) := d

r μ, so μ = rd ν . Now d

r k = m and, since μi |r|dμi , wehave νi |d|rνi for all i. Thus ν ∈ Par(m;d, r). From this we see that the map μ �→ ν(μ) is a bijectionfrom Par(k; r,d) to Par(m;d, r). Thus we have



n−1∑j=0

θrm, jdk t j =

∑ν∈Par(m;d,r)

π rd ν

l(ν)∏i=1

(t A r

d νi−1(t)[d]rd νit

)d,νi

as claimed. �Theorem 5.8. Say λ ∈ Par(n) and n = kd.

(1) If there is some r ∈ [n] such that d does not divide rmr(λ) then θλ, jdk = 0 for all 0 � j � n − 1.

(2) If d divides rmr(λ) for all r ∈ [n] then

n−1∑j=0

θλ, jdk t j =

( nd

1m1(λ)d ,

2m2(λ)d , . . . ,

nmn(λ)d

) n∏r=1

rmr(λ)−1∑j=0

θrmr (λ), jdrmr (λ)/d t j.

Proof. Let σ ∈ Sλ , so σ can be written as a product of disjoint cycles in which there appear exactlymr(λ) r-cycles for each r ∈ [n]. For each such r, let σr be the product of all these mr(λ) r-cycles,and let Br be the support of σr . If σ ∈ C Sn (γ

kn ) then γ k

n commutes with each σr . It follows that γ kn

stabilizes each Br setwise. Therefore, for each r ∈ [n], there is some Yr ⊆ [k] such that

Br =⊎i∈Yr

Xi,

where Xi = {h ∈ [n]: h ≡ i mod k}. Since |Xi| = d for all i, it follows that rmr(λ) = |Br | = d|Yr |, so (1)holds.

For each r ∈ [n], let βr ∈ S Br act as γ kn does on Br . We have γ k

n = ∏nr=1 βr , and βr commutes with

σr for all r. For each r ∈ [n], let fr be the unique order preserving bijection from Br to [rmr(λ)]. Directcalculation shows that for each r, we have

f −1r βr fr = γ

rmr(λ)/drmr(λ) . (50)

Also,

exc(σ ) =n∑

r=1

exc(σr) =n∑

r=1

exc(

f −1r σr fr

). (51)

On the other hand suppose we are given an ordered n-tuple (Y1, . . . , Yn) of subsets of [k] suchthat

(a) |Yr | = rmr(λ)/d for each r ∈ [n], and(b) [k] = ⊎n

r=1 Yr ,

and we set Br = �i∈Yr Xi for each r. Then each Br is γ kn -invariant, and if we set βr equal to the

restriction of γ kn to Br , we can obtain σ ∈ C Sn (γ

kn ) ∩ Sλ by choosing, for each r, any σr ∈ S Br of type

rmr (λ) commuting with βr and setting σ = ∏nr=1 σr . The number of n-tuples satisfying (a) and (b) is

( nd

1m1(λ)d ,

2m2(λ)d , . . . ,

nmn(λ)d

),

and the theorem now follows from (50) and (51). �



Comparing Theorem 5.7 with Theorem 5.2 and then comparing Theorem 5.8 with Theorem 5.3, weobtain

χQ λ, j

dk = θλ, jdk

for all λ ∈ Par(n), j ∈ {0,1, . . . ,n − 1}, and d,k ∈ P such that dk = n. Theorem 1.2 now follows fromProposition 3.2.

6. Some additional results

As mentioned in the introduction, Theorem 1.2 is a refinement of (5). In this section we showthat the less refined result can also be obtained as a consequence of [16, Corollary 4.3], which statesthat

Amaj,exc,fixn (q, t, s) =

� n2 �∑

m=0

∑k0�0

k1,...,km�2∑ki=n

[n

k0, . . . ,km

]q

sk0

m∏i=1

tq[ki − 1]tq. (52)

Although the alternative proof does not directly involve the Eulerian quasisymmetric functions, theproof of (52) given in [16] does. Hence the Eulerian quasisymmetric functions play an indirect role.In this section we also prove the identities (7) and (8) mentioned in the introduction and as a conse-quence of (8) obtain another cyclic sieving result.

Theorem 6.1. Let dk = n. Then the following expressions are all equal.

(i) Amaj,exc,fixn (ωd, tω−1

d , s),

(ii)∑

σ∈C Sn (γ kn )

texc(σ )sfix(σ ) ,

(iii) Aexc,fixk (t, sd+t[d−1]t[d]t

)[d]kt .

Proof. Let us prove first that (ii)=(iii). For ρ ∈ Sk and i ∈ [k] set

fρ,i(t, s) :=⎧⎨⎩

t[d]t if i ∈ Exc(ρ),

sd + t[d − 1]t if i ∈ Fix(ρ),

[d]t otherwise.

It follows from Lemma 5.4 that

∑σ∈Φ−1(ρ)

texc(σ )sfix(σ ) =k∏

i=1

fρ,i(t, s)

= texc(ρ)[d]k−fix(ρ)t

(sd + t[d − 1]t

)fix(ρ).

By summing over all ρ ∈ Sk , we obtain the equality of the expressions in (ii) and (iii).



Now we prove that (i)=(iii). By setting q = 1 in (1) we obtain

1 +∑k�1

Aexc,fixk

(t,

sd + t[d − 1]t

[d]t

)[d]k

tzk

k! = (1 − t)e(sd+t[d−1]t )z

et[d]t z − te[d]t z

= (1 − t)esd z

e(t[d]t−t[d−1]t )z − te([d]t−t[d−1]t )z

= (1 − t)esd z

etd z − tez. (53)

It follows from (52) and Proposition 2.2 that

Amaj,exc,fixdk

(ωd, tω−1

d , s) =

∑m�0

∑l0�0

l1,...,lm�1∑li=k

(k

l0, . . . , lm

)sdl0

m∏i=1

t[dli − 1]t .

Hence, by straightforward manipulation of formal power series we have,

1 +∑k�1

Amaj,exc,fixdk

(ωd, tω−1

d , s) zk

k! =∑k�0

∑m�0

∑l0�0

l1,...,lm�1∑li=k

(k

l0, . . . , lm

)sdl0

m∏i=1

t[dli − 1]tzk

k!

= esd z∑

m,k�0

∑∑

li=kl1,...,lm�1

(k

l1, . . . , lm

) m∏i=1

t[dli − 1]tzk

k!

= esd z∑m�0

∑l1,...,lm�1

m∏i=1

t[dli − 1]tzli

li !

= esd z∑m�0

(∑l�1

t[dl − 1]tzl

l!)m

.

Further manipulation yields,

∑m�0

(∑l�1

t[dl − 1]tzl

l!)m

= 1

1 − (∑

l�1 t[dl − 1]tzl

l! )

= 1 − t

1 − t + ∑l�1 t(tdl−1 − 1) zl

l!

= 1 − t

1 − t + etd z − 1 − t(ez − 1)

= 1 − t

etd z − tez.



Hence

1 +∑k�1

Amaj,exc,fixdk

(ωd, tω−1

d , s) zk

k! = (1 − t)esd z

etd z − tez.

The result now follows from (53). �Corollary 6.2. Let dk = n. Then

Amaj,excn

(ωd, tω−1

d

) = Ak(t)[d]kt .

Similar results hold for cycle-type q-Eulerian polynomials.

Theorem 6.3. Let dk = n. Then

Amaj,exc(n)

(ωd, tω−1

d

) = (t Ak−1(t) [d]k

t

)d (54)

and

Amaj,exc(n+1)

(ωd, tω−1

d

) = t Ak(t) [d]kt . (55)

Proof. To prove (54), we apply the first equation of Proposition 3.2 which tells us that for all j, the

coefficient of t j in Amaj,exc(n) (ωd, tω−1

d ) is equal to χQ (n), j

dk . By Theorem 4.1, χQ (n), j

dk equals the coefficient

of t j in (t Ak−1(t)[d]kt )d , as Gdk (t) = (t Ak−1(t)[d]k

t )d .To prove (55), we apply the second equation of Proposition 3.2 which tells us that for all j, the

coefficient of t j in Amaj,exc(n+1) (ωd, tω−1

d ) is equal to χQ (n), j

1dk . By Theorem 4.1, χQ (n), j

1dk equals the coefficient

of t j in t Ak(t)[d]kt , as G1dk (t) = t Ak(t)[d]k

t . �Corollary 6.4. Let Sn, j be the set of permutations in Sn with j excedances. Then the triple (Gn, Sn, j,

a(n+1), j+1(q)) exhibits the cyclic sieving phenomenon for all j ∈ {0,1, . . . ,n − 1}.

Proof. That the triple exhibits the cyclic sieving phenomenon is equivalent to the equation

t∑

σ∈C Sn (g)

texc(σ ) = Amaj,exc(n+1)

(ωd, tω−1

d

),

for all divisors d of n and g ∈ Gn of order d. This equation is a consequence of Theorems 6.1 and 6.3,

which respectively say that the left side and the right side of the equation both equal t Ak(t)[d]ndt . �

Acknowledgment

We thank an anonymous referee for a very careful reading of the original version of this paper.

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Author's personal copy · 2011. 4. 3. · quasisymmetric functions and other key terms can be found in Section 2. Shareshian and Wachs [15,16] derive a formula for the generating